Problem: Starting at home, Daniel traveled uphill to the hardware store for 20 minutes at just 18 mph. He then traveled back home along the same path downhill at a speed of 36 mph. What is his average speed for the entire trip from home to the hardware store and back?
Solution: The average speed is not just the average of 18 mph and 36 mph. He traveled for a longer time uphill (since he was going slower), so we can estimate that the average speed is closer to 18 mph than 36 mph. To calculate the average speed, we will make use of the following: $\text{average speed} = \dfrac{{\text{total distance}}}{{\text{total time}}}$ $\text{distance uphill} = \text{distance downhill}$ What was the total distance traveled? ${\begin{align*}\text{total distance} &= \text{distance uphill} + \text{distance downhill}\\ &= 2 \times \text{distance uphill}\end{align*}}$ $\begin{align*}\text{distance uphill} &= \text{speed uphill} \times \text{time uphill} \\\ &= 18\text{ mph} \times 20\text{ minutes}\times\dfrac{1 \text{ hour}}{60 \text{ minutes}}\\ &= 6\text{ miles}\end{align*}$ Substituting to find the total distance: ${\text{total distance} = 12\text{ miles}}$ What was the total time spent traveling? ${\text{total time} = \text{time uphill} + \text{time downhill}}$ $\begin{align*}\text{time downhill} &= \dfrac{\text{distance downhill}}{\text{speed downhill}}\\ &= \dfrac{6\text{ miles}}{36\text{ mph}}\times\dfrac{60 \text{ minutes}}{1 \text{ hour}}\\ &= 10\text{ minutes}\end{align*}$ ${\begin{align*}\text{total time} &= 20\text{ minutes} + 10\text{ minutes}\\ &= 30\text{ minutes}\end{align*}}$ Now that we know both the total distance and total time, we can find the average speed. $\begin{align*}\text{average speed} &= \dfrac{{\text{total distance}}}{{\text{total time}}}\\ &= \dfrac{{12\text{ miles}}}{{30\text{ minutes}}}\times\dfrac{60 \text{ minutes}}{1 \text{ hour}}\\ &= 24\text{ mph}\end{align*}$ The average speed is 24 mph, and which is closer to 18 mph than 36 mph as we expected.